# Prove the following by using the principle of mathematical induction for all n ∈ N:

Question:

Prove the following by using the principle of mathematical induction for all $n \in N: 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots+\frac{1}{(1+2+3+\ldots n)}=\frac{2 n}{(n+1)}$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots n}=\frac{2 n}{n+1}$

For $n=1$, we have

$P(1): 1=\frac{2.1}{1+1}=\frac{2}{2}=1$ which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{2 k}{k+1}$    $\ldots$ (i)

We shall now prove that P(k + 1) is true.

Consider

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}+\frac{1}{1+2+3+\ldots+k+(k+1)}$

$=\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots k}\right)+\frac{1}{1+2+3+\ldots+k+(k+1)}$

$=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+k+(k+1)}$   [Using (i)]

$=\frac{2 k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)} \quad\left[1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]$

$=\frac{2 k}{(k+1)}+\frac{2}{(k+1)(k+2)}$

$=\frac{2}{(k+1)}\left(k+\frac{1}{k+2}\right)$

$=\frac{2}{k+1}\left(\frac{k(k+2)+1}{k+2}\right)$

$=\frac{2}{(k+1)}\left(\frac{k^{2}+2 k+1}{k+2}\right)$

$=\frac{2 \cdot(k+1)^{2}}{(k+1)(k+2)}$

$=\frac{2(k+1)}{(k+2)}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.